exercise 1.4b
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@ -229,4 +229,28 @@ Clearly $d_u$ is a metric.
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\begin{claim}
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\begin{claim}
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There exists a countable dense subset.
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There exists a countable dense subset.
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\end{claim}
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\end{claim}
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\todo{handwritten solution}
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\begin{subproof}
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Fix a metric $d_X$ on $X$ defining its topology.
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Let
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\[
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C_{m,n} \coloneqq \{f \in \cC(X,Y) : \forall x,y \in X.~\left( d_X(x,y) < \frac{1}{m+1} \implies d(f(x), f(y)) <\frac{1}{n+1}\right) \}.
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\]
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Choose $X_m \subseteq X$ finite with $X \subseteq \bigcup_{x \in X_m} B_{\frac{1}{m+1}}(x)$.
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Let $D_{m,n} \subseteq C_{m,n}$ be countable,
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such that for every $f \in C_{m,n}$ and every $\eta > 0$,
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there is $g \in D_{m,n}$ with $d(f(y), g(y)) < \frac{\eta}{3}$
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for each $y \in X_m$.
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Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$:
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Indeed if $f \in \cC(X,Y)$ and $\eta > 0$,
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we finde $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$,
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since $f$ is uniformly continuous.
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Let $g \in D_{m,n}$ be such that $\forall y \in X_m.~d(f(y), g(y)) < \frac{1}{n+1}$.
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We have $d_u(f,g) \le \eta$,
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since for every $x \in X$, we find $y \in X_m$ with $d_X(x,y) < \frac{1}{m+1}$,
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hence
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\begin{IEEEeqnarray*}{rCl}
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d_Y(f(x), g(x)) &\le& d_Y(f(x), f(y)) + d_Y(f(y), g(y)) + d_Y(g(y), g(x))\\
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&\le& \frac{1}{n+1} + \frac{1}{n+1} + \frac{1}{n+1} \le \eta.
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\end{IEEEeqnarray*}
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\end{subproof}
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